Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k^2 + 17k + 72}{-2k^2 - 30k - 108} \div \dfrac{k - 4}{-3k - 18} $
Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k^2 + 17k + 72}{-2k^2 - 30k - 108} \times \dfrac{-3k - 18}{k - 4} $ First factor out any common factors. $n = \dfrac{k^2 + 17k + 72}{-2(k^2 + 15k + 54)} \times \dfrac{-3(k + 6)}{k - 4} $ Then factor the quadratic expressions. $n = \dfrac {(k + 9)(k + 8)} {-2(k + 9)(k + 6)} \times \dfrac {-3(k + 6)} {k - 4} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { (k + 9)(k + 8) \times -3(k + 6)} { -2(k + 9)(k + 6) \times (k - 4)} $ $n = \dfrac {-3(k + 9)(k + 8)(k + 6)} {-2(k + 9)(k + 6)(k - 4)} $ Notice that $(k + 9)$ and $(k + 6)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-3\cancel{(k + 9)}(k + 8)(k + 6)} {-2\cancel{(k + 9)}(k + 6)(k - 4)} $ We are dividing by $k + 9$ , so $k + 9 \neq 0$ Therefore, $k \neq -9$ $n = \dfrac {-3\cancel{(k + 9)}(k + 8)\cancel{(k + 6)}} {-2\cancel{(k + 9)}\cancel{(k + 6)}(k - 4)} $ We are dividing by $k + 6$ , so $k + 6 \neq 0$ Therefore, $k \neq -6$ $n = \dfrac {-3(k + 8)} {-2(k - 4)} $ $ n = \dfrac{3(k + 8)}{2(k - 4)}; k \neq -9; k \neq -6 $